How to solve $\frac{dy}{dt}=\alpha y-\beta y^n$ for $n\geq 2,\beta>0$? Is there a way to solve $\frac{dy}{dt}=\alpha y-\beta  y^n$ for $n\geq 2,\beta>0$?
I know how to solve it for $n=2$ but I am not sure how to solve for $n\geq 2$?
 A: Divide by $y^n $ put $\dfrac {1 }{y^{n-1}}=u $ so $\dfrac {1-n}{y^n}.\dfrac {dy}{dt}=\dfrac {du}{dt }$ can you now adjust the original equation to get a first order differential in u. And solve it by known method.
A: $$\text{y}'\left(t\right)=\alpha\cdot\text{y}\left(t\right)-\beta\cdot\text{y}\left(t\right)^\text{n}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(t\right)}{\alpha\cdot\text{y}\left(t\right)-\beta\cdot\text{y}\left(t\right)^\text{n}}\space\text{d}t=\int1\space\text{d}t\tag1$$
Now, use:


*

*Substitute $\text{u}=\text{y}\left(t\right)$ and $\text{d}\text{u}=\text{y}'\left(t\right)\space\text{d}t$:
$$\int\frac{\text{y}'\left(t\right)}{\alpha\cdot\text{y}\left(t\right)-\beta\cdot\text{y}\left(t\right)^\text{n}}\space\text{d}t=\int\frac{1}{\alpha\cdot\text{u}-\beta\cdot\text{u}^\text{n}}\space\text{d}\text{u}=\frac{\ln\left|\alpha\cdot\text{y}\left(t\right)^{1-\text{n}}-\beta\right|}{\alpha\cdot\left(1-\text{n}\right)}+\text{C}_1\tag2$$

*$$\int1\space\text{d}t=t+\text{C}_2\tag3$$


So:
$$\frac{\ln\left|\alpha\cdot\text{y}\left(t\right)^{1-\text{n}}-\beta\right|}{\alpha\cdot\left(1-\text{n}\right)}=t+\text{C}\space\Longleftrightarrow\space\left|\alpha\cdot\text{y}\left(t\right)^{1-\text{n}}-\beta\right|=\text{C}\cdot\exp\left[\alpha\cdot t\cdot\left(1-\text{n}\right)\right]\tag4$$
